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Whitepunk [10]
3 years ago
8

The cart is initially at rest. Force

Physics
1 answer:
Oliga [24]3 years ago
5 0

Answer:

v'=\dfrac{1}{2}v

Explanation:

Given that,

Initial speed of the cart, u = 0

Let F force is applied to the cart for time \Delta t after which the car has speed v. The force on an object is given by :

F = ma

m is the mass of the cart

We need to find the speed of second cart, if the same force is applied for the same time to a second cart with twice the mass. Force becomes,

F=\dfrac{mv'}{t}

v'=\dfrac{F t}{2m}

v'=\dfrac{1}{2}v

So, the speed of second cart is half of the initial speed of first cart. So, the correct option is (b).

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The pilot might be correct (I think), because, if the gravity of the planet is strong, then the planet’s gravity will pull the spaceship into its orbit, so the engines don’t need to be on for the ship to get pushed toward the planet.
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When you look ahead while driving, it is best to:
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I think its [B]
Personally i would say [B] only because If you are looking beyond the car in front of you..... then what if the car in front of you throws on breaks... you would hit them in the butt because you weren't paying attention to the car.
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A rock is thrown upward with a velocity of 22 meters per second from the top of a 25 meter high cliff, and it misses the cliff o
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Answer:

The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.

Explanation:

We will use the equations of motion for this.

u = initial velocity of the rock = 22 m/s

g = acceleration due to gravity = -9.8 m/s²

y = vertical position of the rock at a time t = 9 m

y₀ = initial height of the rock = 25 m

t = time it takes for the rock to reach height of 9 m.

(y-y₀) = ut + 0.5gt²

(9 - 25) = 22t + 0.5(-9.8)t²

- 14 = 22t - 4.9t²

4.9t² - 22t - 14 = 0

solving this quadratic equation,

t = 5.055 s or - 0.565 s

Since time cannot be negative,

t = 5.055 s = 5.06 s

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3 years ago
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Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding. Disregarding friction, what is the velo
lilavasa [31]

Answer:

The velocity of the Mr. miles is 17.14 m/s.

Explanation:

It is given that,

Mr. Miles zips down a water-slide starting at 15 m vertical distance up the scaffolding, h = 15 m

We need to find the velocity of the Mr. Miles at the bottom of the slide. It is a case of conservation of energy which states that the total energy of the system remains conserved. Let v is the velocity of the Mr. miles. So,

v=\sqrt{2gh}

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v=\sqrt{2\times 9.8\times 15}

v = 17.14 m/s

So, the velocity of the Mr. miles is 17.14 m/s. Hence, this is the required solution.

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